[ back to MATH 102/03 home ] | revised 2001-09-28 |

Please note: For most of problems 11 through 24, alternative solution methods are possible. Any mathematically valid method is acceptable.

1. This is false, and many corrections are possible. Some are (3-2), -(2-3), and 1.

2. True.

3. rational (A terminating or repeating decimal can be expressed as a fraction of integers, a rational number. Divide 56 by 9 on your calculator and see what you get. If your exam had 6.333..., try dividing 19 by 3.)

4. | x^{-2} |

3 |

5. is not (The 1/x term is the problem.)

6. 1/x

7. True (You can't "cancel" the 2 or the x, because on the top they are multiplied and on the bottom they are added.)

8. <---+---+---(***+***+***+***+***+***+****> -4 -3 -2 -1 0 1 2 3 4The interval is everything to the right of -2, not including -2.

An open circle instead of the ( above -2 is acceptable. A filled-in circle or a [ is incorrect; that would be right for "greater than or equal to".

9. [-3,5]

(You need a closed interval here. (-3,5) would be
right for the inequality "-3 < x < 5".)

10. x < 4

11. 3x*x²+3x*4 = 3x³+12x

"*" is the raised dot for multiplication.)

12. | |x+2| | |-3+2| | 1 | |||

= | = | = -1 | ||||

-3+2 | -1 | -1 | ||||

Key point here: |x+2| is not the same as |x|+|2|. |

13. (p²)^{-4} = p^{2×(-4)} =
p^{-8} = 1/p^{8}

14. Writing "cbrt" for the cube root, cbrt(p^{6}) =
p^{6/3} = p²

Or, cbrt(p^{6}) = cbrt( [p²]³ ) =
p²

15. yº = 1; therefore 2yº = 2; therefore 4y²(2yº)³ = 4y²(2)³ = 4y²*8 = 32y²

16. 2x²y-16xy+24y = 2y(x²-8x+12) = 2y(x-2)(x-6)

17. See special factor patterns, page 28 of your book. This problem
wants you to recognize that 9z^{4} = (3z²)² and
therefore

9z^{4}-25 = (3z²)²-5² =
(3z²+5)(3z²-5)

18. (a) | x³-8 | (x-2)(x²+2x+4) | |

= | = x²+2x+4 | ||

x-2 | x-2 |

(b) Domain: x =/= 2, or all reals except 2.

1 sqrt(3) sqrt(3) 19. ------- * ------- = ------- sqrt(3) sqrt(3) 3See page 18 of your textbook. (Some students gave an answer of 0.577 or similar. That's approximately correct, but the above is exactly correct. The distinction will be very important when we do Chapter 5.)

2 sqrt(x)+2 2(sqrt(x)+2) 2sqrt(x) + 4 20. --------- * --------- = --------------- = ------------ sqrt(x)-2 sqrt(x)+2 [sqrt(x)]² - 2² x - 4See page 19 of your textbook.

4x-2 2x-5 (4x-2) - (2x-5) 4x - 2 - 2x + 5 2x+3 21. ---- - ---- = --------------- = --------------- = ---- x-3 x+3 x+3 x+3 x+3No further simplification is possible.

4y-16 2y+6 4(y-4) 2(y+3) 4*2 8 22. ----- * ---- = ------ * ------- = ------ = - --- 5y+15 4-y 5(y+3) -1(y-4) 5*(-1) 5Some students chose to multiply (4y-16)(2y+6) by (5y+15)(4-y) and then simplify. While that's mathematically valid, it's more work and it's very easy to make an error that way. If you did it that way and your work was correct, you still got full points.

23. This problem wants you to use the special product (u-v)³
from page 26 of your textbook. Many students chose to multiply out
(2x-3y)(2x-3y)(2x-3y), which is valid but a lot more work and also
error prone. If you multiplied out correctly, you got the full 6
points. The short way is

(2x-3y)³ =

(2x)³ - 3(2x)²(3y) + 3(2x)(3y)² - (3y)³ =

8x³ -3(4x²)(3y) + 3(2x)(9y²) - 27y³ =

8x³ - 36x²y + 54xy² - 27y³

2 2 2(x-1) 2(x+1) 2x-2 + 2x+2 4x 24. --- + --- = ---------- + ---------- = ----------- = ---- x+1 x-1 (x+1)(x-1) (x+1)(x-1) (x+1)(x-1) x²-1

Questions? Ask your instructor or a Baker Center tutor. (MWF 8 am students can also E-mail Stan Brown.)

You will definitely need to know how to do problems like this for the rest of the course.